What is the integral from-2 to-1 of x^2e^{x^3+8} ?

The integral from-2 to-1 of x^2e^{x^3+8} is (e^7-1)/3

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integral from-2 to-1 of x^2e^{x^3+8}

\int_{- 2}^{- 1} x^{2} e^{x^{3} + 8} d x

\frac{e ^{7} - 1}{3}

Decimal Notation

365.21105 \dots

Solution steps

\int_{- 2}^{- 1} x^{2} e^{x^{3} + 8} d x

Apply u-substitution

: \int_{0}^{7} \frac{e ^{u}}{3} d u

= \int_{0}^{7} \frac{e ^{u}}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int_{0}^{7} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{3} [e^{u}]_{0}^{7}

Compute the boundaries:

e^{7} - 1

= \frac{1}{3} (e^{7} - 1)

Simplify

= \frac{e ^{7} - 1}{3}

9 π \cdot R^{2}

f (x) = \frac{sec ( x )}{x}

\frac{\partial}{\partial x} (6 x y^{5} + 4 y^{3})

\frac{d}{d x} (- x^{π})

\int \frac{x + 3}{1 - x ^{2}} d x

What is the integral from-2 to-1 of x^2e^{x^3+8} ?
The integral from-2 to-1 of x^2e^{x^3+8} is (e^7-1)/3